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\title{\heiti\zihao{2} 习题15.3}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求下列函数的二阶偏导数}
\subsection{$z=\tan \dfrac{x}{y}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x}             & =\dfrac{1}{y\cos^2\dfrac{x}{y}}                          \\
		\dfrac{\partial z}{\partial y}             & =-\dfrac{x}{y^2\cos^2\dfrac{x}{y}}                       \\
		\dfrac{\partial^2 z}{\partial x^2}         & =\dfrac{2\sin \dfrac{x}{y}}{y^2\cos^3\dfrac{x}{y}}       \\
		\dfrac{\partial^2 z}{\partial y^2}         & =\dfrac{2x(x\tan \dfrac{x}{y}+y)}{y^4\cos^2\dfrac{x}{y}} \\
		\dfrac{\partial^2 z}{\partial y\partial x} & =-\dfrac{y+2x\tan \dfrac{x}{y}}{y^3\cos^2 \dfrac{x}{y}}  \\
		\dfrac{\partial^2 z}{\partial x\partial y} & =-\dfrac{y+2x\tan \dfrac{2x}{y}}{y^3\cos^2 \dfrac{x}{y}} \\
	\end{aligned}
$$

\subsection{$u=\mathrm{e}^{x y z}$}
\textbf{解}\quad
由于$u$的二阶导数均存在且在$\mathbb{R}^2$上连续,从而
$$
	\begin{array}{ccc}
		\dfrac{\partial u}{\partial x} = yz\mathrm{e}^{xyz}                   & \dfrac{\partial u}{\partial y} = xz\mathrm{e}^{xyz}                   &
		\dfrac{\partial u}{\partial z} = xy\mathrm{e}^{xyz}                                                                                             \\
		\dfrac{\partial^2 u}{\partial x^2} = y^2z^2\mathrm{e}^{xyz}           & \dfrac{\partial^2 u}{\partial y^2} = x^2z^2\mathrm{e}^{xyz}           &
		\dfrac{\partial^2 u}{\partial z^2} = x^2y^2\mathrm{e}^{xyz}                                                                                     \\
		\dfrac{\partial^2 u}{\partial y\partial x} = z\mathrm{e}^{xyz}(1+xyz) & \dfrac{\partial z\partial y}{\partial y^2} = x\mathrm{e}^{xyz}(1+xyz) &
		\dfrac{\partial^2 u}{\partial x\partial z} = y\mathrm{e}^{xyz}(1+xyz)                                                                           \\
	\end{array}
$$

\subsection{$z=f\left(x, \dfrac{x}{y}\right)$}
\textbf{解}\quad
$$
	\begin{array}{cc}
		\dfrac{\partial z}{\partial x} = f_1 + f_2\dfrac{1}{y}                                                     & \dfrac{\partial z}{\partial y} = -f_2\dfrac{x}{y^2}                                                     \\
		\dfrac{\partial^2 z}{\partial x^2} = f_{11} + \dfrac{1}{y}f_{12} + f_{21}\dfrac{1}{y}+f_{22}\dfrac{1}{y^2} & \dfrac{\partial^2 z}{\partial y^2} = \dfrac{2x}{y^3}f_2+\dfrac{x^2}{y^4}f_{22}                          \\
		\dfrac{\partial^2 z}{\partial y\partial x} = -\dfrac{1}{y^2}\left(f_{22}\dfrac{x}{y}+f_{12}x+f_2\right)     & \dfrac{\partial^2 z}{\partial x\partial y} = -\dfrac{1}{y^2}\left(f_2+f_{21}x+f_{22}\dfrac{x}{y}\right)
	\end{array}
$$
\subsection{$u=\dfrac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}$}
\textbf{解}\quad
由于$u$的二阶导数均存在且在$\mathbb{R}^{3+}$上连续,从而
$$
	\begin{array}{ccc}
		\dfrac{\partial u}{\partial x} = -\dfrac{x}{\left(x^2+y^2+z^2\right)^{3/2}}      & \dfrac{\partial u}{\partial y} = -\dfrac{y}{\left(x^2+y^2+z^2\right)^{3/2}}     & \dfrac{\partial u}{\partial z} = -\dfrac{z}{\left(x^2+y^2+z^2\right)^{3/2}}     \\
		\dfrac{\partial^2 u}{\partial x^2} = -\dfrac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^{5/2}} & \dfrac{\partial^2 u}{\partial y^2} = -\dfrac{x^2-2y^2+z^2}{(x^2+y^2+z^2)^{5/2}} & \dfrac{\partial^2 u}{\partial z^2} = -\dfrac{x^2+y^2-2z^2}{(x^2+y^2+z^2)^{5/2}} \\
		u_{yx} = u_{xy} = \dfrac{3xy}{(x^2+y^2+z^2)^{5/2}}                               & u_{zy} = u_{yz} = \dfrac{3yz}{(x^2+y^2+z^2)^{5/2}}                              & u_{xz} = u_{zx} = \dfrac{3zx}{(x^2+y^2+z^2)^{5/2}}
	\end{array}
$$
\subsection{$u=\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}$}
\textbf{解}\quad
由于$u$的二阶导数均存在且在$\mathbb{R}^{n+}$上连续,从而
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x_{i}}                 & =\dfrac{x_i}{\sqrt{\sum\limits_{k=1}^{n}x_k^2}}                                            \\
		\dfrac{\partial^2 u}{\partial x_{i}^2}             & =\dfrac{\sum\limits_{k=1,k\neq i}^{n}x_k^2}{\left(\sum\limits_{k=1}^{n}x_k^2\right)^{3/2}} \\
		\dfrac{\partial^2 u}{\partial x_{j}\partial x_{i}} & =\dfrac{-x_ix_j}{\left(\sum\limits_{k=1}^{n}x_k^2\right)^{3/2}}
	\end{aligned}
$$
\subsection{$u=\ln \left(x_{1}+x_{2}+\cdots+x_{n}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x_{i}}                 & =\dfrac{1}{\sum\limits_{k=1}^{n}x_k}                 \\
		\dfrac{\partial^2 u}{\partial x_{i}^2}             & =-\dfrac{1}{\left(\sum\limits_{k=1}^{n}x_k\right)^2} \\
		\dfrac{\partial^2 u}{\partial x_{j}\partial x_{i}} & =-\dfrac{1}{\left(\sum\limits_{k=1}^{n}x_k\right)^2}
	\end{aligned}
$$

\section{验证函数 $z=\ln \sqrt{x^{2}+y^{2}}$ 满足方程 $\dfrac{\partial^{2} z}{\partial x^{2}}+\dfrac{\partial^{2} z}{\partial y^{2}}=0$.}
\begin{proof}
	$$
		\begin{aligned}
			\dfrac{\partial^2 z}{\partial x^2} & = \dfrac{-x^2+y^2}{(x^2+y^2)^2} \\
			\dfrac{\partial^2 z}{\partial y^2} & = \dfrac{x^2-y^2}{(x^2+y^2)^2}
		\end{aligned}
	$$
	显然
	$$
		\dfrac{\partial^{2} z}{\partial x^{2}}+\dfrac{\partial^{2} z}{\partial y^{2}}=0
	$$
\end{proof}

\section{设 $y=\varphi(x+a t)+\psi(x-a t)$, 验证 $\dfrac{\partial^{2} y}{\partial t^{2}}=a^{2} \dfrac{\partial^{2} y}{\partial x^{2}}$.}
\begin{proof}
	$$
		\begin{aligned}
			\dfrac{\partial y}{\partial t}         & =a \varphi^{\prime}(x+a t)-a \psi^{\prime}(x-a t)                       \\
			\dfrac{\partial^{2} y}{\partial t^{2}} & =a^{2} \varphi^{\prime \prime}(x+a t)+a^{2} \psi^{\prime \prime}(x-a t) \\
			\dfrac{\partial y}{\partial x}         & =\varphi^{\prime}(x+a t)+\psi^{\prime}(x-a t)                           \\
			\dfrac{\partial^{2} y}{\partial x^{2}} & =\varphi^{\prime \prime}(x+a t)+\psi^{\prime \prime}(x-a t)
		\end{aligned}
	$$
	故
	$$
		\dfrac{\partial^{2} y}{\partial t^{2}}=a^{2} \dfrac{\partial^{2} y}{\partial x^{2}}
	$$
\end{proof}
\section{设 $f, x, y$ 都是具有二阶连续偏导数的二元函数,记 \\$u=f(x(s, t), y(s, t))$, 求 $\dfrac{\partial^{2} u}{\partial s^{2}}$, $\dfrac{\partial^{2} u}{\partial t^{2}}, \dfrac{\partial^{2} u}{\partial s \partial t}$.}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial^{2} u}{\partial s^{2}}        & =\left(f_{11}\dfrac{\partial x}{\partial s}+f_{12}\dfrac{\partial y}{\partial s}\right)\dfrac{\partial x}{\partial s}+\left(f_{21}\dfrac{\partial x}{\partial s}+f_{22}\dfrac{\partial y}{\partial s}\right)\dfrac{\partial y}{\partial s}+f_1\dfrac{\partial^2 x}{\partial s^2}+f_2\dfrac{\partial^2 y}{\partial s^2}                 \\
		\dfrac{\partial^{2} u}{\partial t^{2}}        & =\left(f_{11}\dfrac{\partial x}{\partial t}+f_{12}\dfrac{\partial y}{\partial t}\right)\dfrac{\partial x}{\partial t}+\left(f_{21}\dfrac{\partial x}{\partial t}+f_{22}\dfrac{\partial y}{\partial t}\right)\dfrac{\partial y}{\partial t}+f_1\dfrac{\partial^2 x}{\partial t^2}+f_2\dfrac{\partial^2 y}{\partial t^2}                 \\
		\dfrac{\partial^{2} u}{\partial s \partial t} & =\left(f_{11}\dfrac{\partial x}{\partial t}+f_{12}\dfrac{\partial y}{\partial t}\right)\dfrac{\partial x}{\partial s}+\left(f_{21}\dfrac{\partial x}{\partial t}+f_{22}\dfrac{\partial y}{\partial t}\right)\dfrac{\partial y}{\partial s}+f_1\dfrac{\partial^2 x}{\partial s\partial t}+f_2\dfrac{\partial^2 y}{\partial s\partial t}
	\end{aligned}
$$

\section{求下列函数在指定点的 Peano 余项 Taylor 公式}
\subsection{$f(x, y)=2 x^{2}-x y-y^{2}-6 x-3 y+5$ 在点 $(1,-2)$ 展到 $n$ 阶}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial f}{\partial x}             & =4x-y-6=0                                                                                            \\
		\dfrac{\partial f}{\partial y}             & =-x-2y-3=0                                                                                           \\
		\dfrac{\partial^2 f}{\partial x^2}         & =4                                                                                                   \\
		\dfrac{\partial^2 f}{\partial y^2}         & =-2                                                                                                  \\
		\dfrac{\partial^2 f}{\partial y\partial x} & =\dfrac{\partial^2 f}{\partial x\partial y}=-1                                                       \\
		\therefore f(x,y)                          & = f(1,-2)+f_xx+f_yy+\dfrac{1}{2}\left(f_{xx}(x-1)^2+2f_{xy}(x-1)(y+2)+f_{yy}(y+2)^2\right)+o(\rho^n) \\
		                                           & =2{\left(x-1\right)}^2-\left(x-1\right)\left(y+2\right)-{\left(y+2\right)}^2+5
	\end{aligned}
$$


\subsection{$f(x, y)=\mathrm{e}^{(x+y)}$ 在点 $(0,0)$ 展到 $n$ 阶}
\textbf{解}\quad
换元,令$x+y=u$,则
$$
	\mathrm{e}^u = \sum_{n=0}^{n}\dfrac{(x+y)^n}{n!}+o(\rho^n)
$$

\subsection{$f(x, y)=\sin x \ln (1+y)$ 在点 $(0,0)$ 展到 $3$ 阶}
\textbf{解}\quad
$$
	\begin{aligned}
		\sin x           & = x + O(x^3)                   \\
		\ln (1+y)        & =y-\dfrac{1}{2}y^2+o(y^2)      \\
		\sin x \ln (1+y) & =xy-\dfrac{1}{2}xy^2+o(\rho^3)
	\end{aligned}
$$
\subsection{$f(x, y)=\sin \left(x^{2}+y^{2}\right)$ 在点 $(0,0)$ 展到 $4$ 阶}
\textbf{解}\quad
换元,令$x^2+y^2=u$,则相当于对$u$展开到二次.
$$
	\sin ua = x^2+y^2+o(\rho^4)
$$

\end{document}